If the principal stresses at a point in an elastic material are 2x (tensile), x (tensile) and \(\frac{x}{2}\) (compressive). What will be the value of 'x', if the material fails according to shear strain energy theory [ Mises and Henkey's theory]? The elastic limit in simple tension is 200 N/mm2.

Concept:

According to **shear strain energy theory (Von Mises and Hencky’s theory)**, the failure criterion is based on the distortion energy. The equivalent stress is given by:

\( \sigma_{eq} = \sqrt{\frac{ (\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}{2}} \)

where:

• \(\sigma_1 = 2x ~ (tensile)\)
• \(\sigma_2 = x ~(tensile)\)
• \( \sigma_3 = \frac{x}{2} ~ (compressive)\)

The material will fail when the equivalent stress equals the **elastic limit stress**:

\( \sigma_{eq} = \sigma_y \)

where \(\sigma_y = 200\) N/mm².

Calculation:

Step 1: Compute equivalent stress

\( \sigma_{eq} = \sqrt{\frac{(2x - x)^2 + (x - (-x/2))^2 + (-x/2 - 2x)^2}{2}} \)

\( \sigma_{eq} = \sqrt{\frac{(x)^2 + (3x/2)^2 + (5x/2)^2}{2}} \)

\( \sigma_{eq} = \sqrt{\frac{x^2 + 9x^2/4 + 25x^2/4}{2}} \)

\( \sigma_{eq} = \sqrt{\frac{4x^2 + 9x^2 + 25x^2}{8}} \)

\( \sigma_{eq} = \sqrt{\frac{38x^2}{8}} = \sqrt{\frac{19x^2}{4}} = \frac{x\sqrt{19}}{2} \)

Step 2: Equating to Elastic Limit

\( \frac{x\sqrt{19}}{2} = 200 \)

\( x = \frac{400}{\sqrt{19}} \)